Question

The tank in Figure P14.15 is filled with water of depth $d=2.00m$. At the bottom of one sidewall is a rectangular hatch of height $h=1.00m$and width $w=2.00m$that is hinged at the top of the hatch. (a) Determine the magnitude of the force the water exerts on the hatch. (b) Find the magnitude of the torque exerted by the water about the hinges.

Short answer

(a) The magnitude of the force the water exerts on the hatch is$F=29.4kN\left(totheright\right)$ .

(b) The magnitude of the torque exerted by the water about the hinges is $\tau =16.3kN·m\left(counterclockwise\right)$.

Step-by-step solution

Pressure:

The pressure in a fluid is the force per unit area exerted by the fluid on a surface:

$P=\frac{F}{A}$

Where P can also be given as:

$P=\rho gh$

Where, P is the pressure, A is area, and F is the force exerted on the fluid.

(a) Determine the magnitude of the force the water exerts on the hatch

The air outside and the water inside both exert atmospheric pressure, so only the excess water pressure $\rho gh$ counts for the net force. Take a strip of hatch between depth h and$h+dh$ . It feels force

$$\begin{gathered} dF=PdA \\ =\rho gh\left(2.00m\right)dh \end{gathered}$$

The total force is defined by,

$$\begin{gathered} F=\int dF \\ =\underset{h=1}{\overset{h=2}{\int }}\rho gh\left(2.0\text{0 m}\right)dh \\ ={\left.\rho gh\left(2.0\text{0 m}\right)\frac{h^2}{2}\right|}_{h=1}^{h=2} \end{gathered}$$

$$\begin{gathered} F=\left(1000Kg/m^3\right)\left(9.8m/s^2\right)\left[\frac{{\left(2.00m\right)}^2-{\left(1.00m\right)}^2}{2}\right]\times 2.00m \\ =29.4Kn(totheright) \end{gathered}$$

(b) Determine the magnitude of the torque exerted by the water around the hinges

The lever arm of df is the distance $\left(h-1.00m\right)$from hinge to strip is,

$$\begin{gathered} \tau =\int d\tau \\ =\underset{h=1}{\overset{h=2}{\int }}\rho gh\left(2.00m\right)\left(h-1.00m\right)dh \\ =\rho g\left(2.00m\right){\left[\frac{h^3}{3}-\left(1.00m\right)\frac{h^2}{2}\right]}_{h=1}^{h=2} \end{gathered}$$

role="math" localid="1663678140182" $$\begin{gathered} \tau =\left(1000Kg/m^3\right)\left(9.8m/s^2\right)\left(2.00m\right)\left[\frac{7.00m^3}{3}-\frac{3.00m^3}{2}\right] \\ =16.3KN.m\left(counterclockwise\right) \end{gathered}$$